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Is $\dfrac{\sin(x)}{x}$ at $x = 0$ continuous? Whats the value at $x=0~?$

Harsh Kumar
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Zeref Dragneel
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  • I believe You do know about something Called L'hospital Rule? – The Dead Legend Apr 13 '17 at 05:22
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    @BrevanEllefsen Why can't you use Hospital? This CAN be approached with Hospital – DMH16 Apr 13 '17 at 05:25
  • This might be helpful http://math.stackexchange.com/questions/2092824/confusion-around-lim-x-to-0-frac-sin-xx-sin-x-x-sin-0-0 – mlc Apr 13 '17 at 05:26
  • @DMH16 L'Hopital assumes you know the derivatives of the top and bottom. However, proving that $\frac{d}{dx} \sin(x) = \cos(x)$ usually USES the limit $\lim_{x \to 0} \frac{\sin x}{x}$ and so the logic is circular. depends on how you define the sine function though. Defining it in terms of power functions let's you do this. – Brevan Ellefsen Apr 13 '17 at 05:27
  • @BrevanEllefsen That is true, however, it wouldn't be wrong to state that the OP can use Hospital to evaluate the limit (if he has been properly introduced to it). You clearly stated that "it doesn't work here", which is incorrect – DMH16 Apr 13 '17 at 05:30
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    @DMH16 It is absolutely wrong unless he can justify his logic. Does L'Hopital apply here? Yes. Is applying it circular logic without some proof of $\frac{d}{dx} \sin x = \cos x$? Absolutely. Therefore, using L'Hopital here might give you the right answer because the conditions are valid, but this is not a proof - it is circular logic. – Brevan Ellefsen Apr 13 '17 at 05:34
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    @DMH16 If you are taking a heuristic approach or simply accepting that we know the derivative of $\sin(x)$ then I suppose you can use L'Hopital, but this isn't a very rigorous approach. – Brevan Ellefsen Apr 13 '17 at 05:36
  • @BrevanEllefsen Generally Hospital's rule is introduced with differentiation, therefore, I assumed it would be correct applying the definition of such rule to this case. If the OP hasn't been introduced to differentiation then the geometric proof should be looked at. However, I insist in underlining that saying "Hospital rule here doesn't work" is a false statement. – DMH16 Apr 13 '17 at 05:40

2 Answers2

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It isn't if you don't define its value at $0$ to be $1$. As it is written, the function $f$, defined by $f(x) = (\sin x)/x$, is not defined at $0$, so it is not continuous at $0$.

However, the function $g$, defined by $$ g(x) = \begin{cases} \dfrac{\sin x}{x} &\text{if $x \neq 0$,}\\[1ex] 1 &\text{if $x = 0$,} \end{cases} $$ is continuous at $0$. Why? Well, this requires you to show that $\lim_{x \to 0} (\sin x)/x = 1$, which I will omit discussing (you can use L'Hospital's rule to handle it, although most calculus textbooks prove it using a geometric argument, since L'Hospital's rule is unavailable at that time to the reader).

Mark Twain
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Note that $\sin x = \sum_{k=0}^\infty {x^{2k+1} \over (2k+1) }$, so for $x \neq 0$, we have ${\sin x \over x} = \sum_{k=0}^\infty {x^{2k} \over (2k+1) }$, and since $\sigma(x) = \sum_{k=0}^\infty {x^{2k} \over (2k+1) }$ is analytic everywhere, we see that $\lim_{x \to 0} {\sin x \over x} = \lim_{x \to 0} \sigma(x) = \sigma(0) = 1$.

The function $\sigma$ is also known as $\operatorname{sinc}$.

copper.hat
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